The specific heat of ice is 0.5 calories/gram.
60 grams of ice will require calories to raise the temperature 1°c.
o
о 30
o
о 60
о120
o
о
0.5

The answer is: 30 calories

From the statement we have:

We can calculate heat transferred using the following equation;

q=mCΔT

Where:

q=HeatTransferred

m=mass=60g

C=SpecificHeat=0.5\frac{calories}{g}

ΔT=TemperatureChange=1°C

So, by substituting we have:

So, 60 grams of ice will require 30 calories to raise the temperature 1°C

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